The Impact of R5 on Req Calculation

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Homework Help Overview

The discussion revolves around the impact of adding a resistor (R5) on the equivalent resistance (Req) calculation in a circuit involving a capacitor. Participants explore how the presence of R5 affects the behavior of other resistors (R2 and R3) that are initially shorted by the capacitor.

Discussion Character

  • Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the effect of R5 on the circuit, questioning how it changes the status of R2 and R3 from being shorted to being included in the Req calculation. There are inquiries about how to redraw the circuit with the capacitor represented as a short circuit.

Discussion Status

Some participants have provided guidance on redrawing the circuit and understanding the capacitor's behavior at time t=0. There is an acknowledgment of the need to explain the capacitor's behavior under specific conditions, indicating a productive exploration of the topic.

Contextual Notes

Participants mention the assumption that the capacitor is uncharged at time t=0, which influences its representation in the circuit. There is also a note on the complexity of the problem if the capacitor had an initial charge or if the current at a different time instant were to be considered.

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Homework Statement
Consider the circuit to the right, with R5 = 119 Ω in series with the capacitor. Once again, the switch has been open for a long time when at time t = 0, the switch is closed. What is I1(0), the magnitude of the current through the resistor R1 just after the switch is closed?
Relevant Equations
I=V/R , Req=R235+R14
So what I know is that without the edition of the R5 resistor R2 and R3 are 0 because the capacitor is short circuiting them. Why does the addition of the R5 resistor cause R2 and R3 to be calculated into the Req for the current equation?
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Adding R5 makes a difference because R2 and R3 are no longer shorted. Redraw the circuit with the capacitor as a "short" to find the initial Req and hence the initial current through R1 the usual way.
 
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kuruman said:
Adding R5 makes a difference because R2 and R3 are no longer shorted. Redraw the circuit with the capacitor as a "short" to find the initial Req and hence the initial current through R1 the usual way.

Thank you I kind of understand that, but how do I redraw the circuit with the capacitor as a "short". I don't really understand what that looks like
 
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mabdeljaber said:
Thank you I kind of understand that, but how do I redraw the circuit with the capacitor as a "short". I don't really understand what that looks like
Replace the capacitor with a wire.
 
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gneill said:
Replace the capacitor with a wire.

Thank you! Understood
 
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Maybe we have to explain why the capacitor at time t=0 behaves as a short. The reason is that the capacitor is assumed to be uncharged so its voltage is zero at this time instant. So if you apply KVL for the time instant t=0 you ll see that the resulting equation is like that the capacitor is a short.

If the capacitor had some initial charge or if we were asked for the current at another time instant then you shouldn't replace the capacitor with a short. You will had to solve differential equations then (unless the charge of the capacitor at that time instant was given).
 

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